Question

For the following separable equations, carry out the indicated analysis. a. Find the general solution of the equation. b. Find the value of the arbitrary constant associated with each initial condition. (Each initial condition requires a different constant.) c. Use the graph of the general solution that is provided to sketch the solution curve for each initial condition. $$y^{2} y^{\prime}(t)=t^{2}+\frac{2}{3} t ; y(-1)=1, y(1)=0, y(-1)=-1$$

Short answer

Question: Determine the solution curves for the differential equation \(y^{\prime}(t)=\frac{t^2+\frac{2}{3}t}{y^{2}}\) with initial conditions \(y(-1)=1, y(1)=0, y(-1)=-1\). Answer: The solution curves corresponding to each initial condition are: a) For \(y(-1)=1\): $$\frac{1}{3}y^3 = \frac{1}{4}t^4 +\frac{1}{9}t^3+\frac{10}{36}$$ b) For \(y(1)=0\): $$\frac{1}{3}y^3 = \frac{1}{4}t^4 +\frac{1}{9}t^3-\frac{13}{36}$$ c) For \(y(-1)=-1\): $$\frac{1}{3}y^3 = \frac{1}{4}t^4 +\frac{1}{9}t^3+\frac{14}{36}$$

Step-by-step solution

Rewrite the equation and separate the variables

Divide both squared terms by \(y^2\) so that we can separate variables: $$y^{\prime}(t)=\frac{t^2+\frac{2}{3}t}{y^{2}}$$ Now, separate the variables: $$y^2 dy = (t^2+\frac{2}{3}t) dt$$

Integrate both sides

Integrating both sides with respect to their respective variables, we have: $$\int y^2 dy = \int (t^2+\frac{2}{3}t) dt$$ Integration yields: $$\frac{1}{3}y^3 = \frac{1}{4}t^4 +\frac{1}{9}t^3+C$$ Where C is the arbitrary constant.

Solve for the arbitrary constant using the initial conditions

We are given 3 initial conditions: \(y(-1)=1, y(1)=0,y(-1)=-1\). We will plug in these values one by one to solve for the arbitrary constants. a) For \(y(-1)=1\): $$\frac{1}{3}(1)^3 = \frac{1}{4}(-1)^4 +\frac{1}{9}(-1)^3+C$$ Solving for C gives: $$C = \frac{10}{36}$$ b) For \(y(1)=0\): $$0 = \frac{1}{4}(1)^4 +\frac{1}{9}(1)^3+C$$ Solving for C gives: $$C = -\frac{13}{36}$$ c) For \(y(-1)=-1\): $$\frac{1}{3}(-1)^3 = \frac{1}{4}(-1)^4 +\frac{1}{9}(-1)^3+C$$ Solving for C gives: $$C = \frac{14}{36}$$

Write the solution curves for each initial condition

Using the values of C determined in Step 3, we have three solutions corresponding to each initial condition: a) For \(y(-1)=1\): $$\frac{1}{3}y^3 = \frac{1}{4}t^4 +\frac{1}{9}t^3+\frac{10}{36}$$ b) For \(y(1)=0\): $$\frac{1}{3}y^3 = \frac{1}{4}t^4 +\frac{1}{9}t^3-\frac{13}{36}$$ c) For \(y(-1)=-1\): $$\frac{1}{3}y^3 = \frac{1}{4}t^4 +\frac{1}{9}t^3+\frac{14}{36}$$ These are the solution curves corresponding to each initial condition.

Key concepts

Initial Conditions

When you solve a differential equation, initial conditions are specific values that allow you to find a particular solution from the general solution. Often, you have a family of solutions, but initial conditions help you single out the one solution that fits these conditions. In this exercise, we have three initial conditions given:

• \( y(-1) = 1 \)
• \( y(1) = 0 \)
• \( y(-1) = -1 \)

Each of these initial conditions influences the value of the arbitrary constant \( C \) in the general solution. By substituting the values of \( y \) and \( t \) from these initial conditions into the integrated equation, we can solve for the corresponding \( C \). Thus, even though we start with a general solution, each initial condition helps us tailor the equation to a specific scenario.

General Solution

In solving differential equations, the general solution represents the collection of all possible solutions to the equation. It typically contains one or more arbitrary constants like \( C \), depending on the order of the differential equation. This exercise demonstrates how to find the general solution of a separable differential equation.
The first step involves rearranging and separating the variables, which enables the integration process. The general solution derived in this exercise is:\[ \frac{1}{3}y^3 = \frac{1}{4}t^4 + \frac{1}{9}t^3 + C \]Here, the function \( y \) is expressed in terms of \( t \), and the solution includes an arbitrary constant \( C \). This equation represents a family of curves, each corresponding to different values of \( C \). This is where integration and initial conditions come into play to find specific solutions.

Integration

Integration is a fundamental part of solving separable differential equations. It allows us to find the antiderivative or the integral of each side of a separated equation. In this exercise, once the equation is separated by variables, integration is applied:

• For the left side: \( \int y^2 \, dy \)
• For the right side: \( \int (t^2 + \frac{2}{3}t) \, dt \)

The integration results in the equation:\[ \frac{1}{3}y^3 = \frac{1}{4}t^4 + \frac{1}{9}t^3 + C \]Notice how each side is integrated independently. This process transforms a differential equation into an algebraic equation, making it easier to work with to find solutions. Different initial conditions can be applied to this general solution through integration constants.

Arbitrary Constant

The arbitrary constant arises in the integration process of differential equations. Every integration introduces an unknown constant \( C \), reflecting the general nature of indefinite integrals.
For each separable differential equation, you will typically have one \( C \) per integral, which can be determined using initial conditions. In this exercise, the constant \( C \) was evaluated differently for each given initial condition:

• For \( y(-1) = 1 \), \( C = \frac{10}{36} \).
• For \( y(1) = 0 \), \( C = -\frac{13}{36} \).
• For \( y(-1) = -1 \), \( C = \frac{14}{36} \).

These calculations offer specific solutions from the general solution by modifying \( C \) based on initial values. Thus, \( C \) plays a crucial role in adapting the general solution to fit real-world conditions or specific scenarios.